Prove that if derivative of a holomorphic function is non-zero, then it is locally univalent Let $f$ be a non-constant entire function such that $f'(z)≠ 0$. Then $f$ is locally univalent.

A proof "by complex analysis", no IFT needed:

Normalize $f$ so $f(0)=0$, $f'(0)=1$: $$f(z)=z+O(|z|^2).$$You can use Rouche to prove

> > There exist $r,\delta>0$ such that if $|\alpha|<\delta$ then $f-\alpha$ has exactly one zero in $\\{|z|
Now choose $\rho\in(0,r)$ so that $|z|<\rho$ implies $|f(z)|<\delta$ and it follows that $f$ is univalent in $\\{|z|<\rho\\}$.